Let f(x) = \frac{x^4 + 2x^3 + 3x^2 + 2x + 1}{x}. Find the minimum value of x for x > 0.
f(x)=x4+2x3+3x2+2x+1x.
Simplify as
f(x) = x^3 + 2x^2 + 3x + 2 + x^(-1) take the derivative and set to 0
3x^2 + 4x + 3 - x^(-2) = 0 multiply through by x^2
3x^4 + 4x^3 + 3x^2 - 1 = 0
Graphing this shows a local min at x ≈ .43426
+1208 
